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  • Solve! - If a curve passes through the pointand has slope of the tangent at any pointon it asthen the c - Limit , continuity and differentiability - JEE Main

If a curve passes through the point \left ( 1,-2 \right ) and has slope of the tangent at any point \left ( x,y \right ) on it as \frac{x^{2}-2y}{x}, then the curve also passes through the point : 

  • Option 1)

    \left ( \sqrt{3},0 \right )

     

     

     

  • Option 2)

    \left ( 3,0 \right )

  • Option 3)

    \left (-1,2 \right )

  • Option 4)

    \left (-\sqrt{2},1 \right )

Answers (1)

best_answer

 

Equation of the tangent -

To find the equation of the tangent we need either one slope + one point or two points.

\therefore \:\:(y-y_{\circ})=m(x_{\circ }-y_{\circ })
 

or\:\:(y-y_{2})=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{2})

- wherein

Where  (x_{\circ},y_{\circ})   is the point on the curve and M = MT  slope of tangent.

 

 

Geometrical interpretation of dy / dx -

Slope of tangent line is tan\theta  where  \theta is the angle made by the line with the  +ve direction of  x  axis.

\therefore \:\:\frac{dy}{dx}=tan\theta

-

\fn_cm \frac{\mathrm{d} y}{\mathrm{d} x}+\frac{2y}{x}=x\\\\If \: \: e^{\int \frac{2}{x}dx}=e^{2lnx}=x^{2}\\\\y(x^{2})=\int x.x^{2}dx\\\\x^{2}y=\frac{x^{4}}{4}+c\\\\\therefore \: y(1)=-2\\\\\Rightarrow c=-\frac{9}{4}\\\\y=\frac{x^{4}-9}{4x^{2}} which\: passes\: through \: (\sqrt{3},0)

 

 


Option 1)

\left ( \sqrt{3},0 \right )

 

 

 

Option 2)

\left ( 3,0 \right )

Option 3)

\left (-1,2 \right )

Option 4)

\left (-\sqrt{2},1 \right )

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